Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 1 (Q.No. 2)
2.
An ammeter of 0-25 A range has a guaranteed accuracy of 1% of full scale reading. The current measured is 5 A. The limiting error is
Answer: Option
Explanation:
.
Discussion:
12 comments Page 1 of 2.
Mangai said:
8 years ago
Limiting error = accuracy * full scale reading = (1%) * 25 = (1/100) *25 = 0.25 A
error in percentage.
% error=(maximum error/scale reading) * 100 = (0.25/ 5) *100 = 5%.
error in percentage.
% error=(maximum error/scale reading) * 100 = (0.25/ 5) *100 = 5%.
(3)
Imran said:
1 decade ago
0.25/5 = 0.05x100 = 5%.
Anil said:
1 decade ago
Please correct it.
% error = 0.25*100/5volt = 5%.
% error = 0.25*100/5volt = 5%.
Sanju said:
1 decade ago
Scale given 1% so 25x1/100 = 0.25A.
%error = 0.25/58x100 = 5%.
%error = 0.25/58x100 = 5%.
Shinykiran said:
10 years ago
How the value 58 came?
Sharada said:
9 years ago
What is the formula used here? Can anyone say?
Bindu said:
8 years ago
I am not understanding. Can anyone explain properly?
Rajeev said:
8 years ago
Here, (% error at desired scale * desired scale ) = (%error at full scale value * full scale value).
Mahesh Thapa Magar said:
8 years ago
%Error(DS) * DS = %Error(FS) * FS.
DS:Desired scale.
FS:Full SCale.
DS:Desired scale.
FS:Full SCale.
Narender said:
8 years ago
1/25*100=4,
4/100*5=20/100,
= 5%.
4/100*5=20/100,
= 5%.
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